How many atoms are in 0.065 grams of Cadmium?

Write the symbol for element x which has 22 electrons and 22 neutrons

marissa [1.9K]

Titanium has 22 protons and electrons. The symbol for Titanium is (Ti).

7 0

3 years ago

The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.

ASHA 777 [7]

We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:

PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas

Hope this answers the question. Have a nice day.

7 0

3 years ago

Differentiate between the three types of simple permanent tissues on the basis of their cell walls

True [87]

Answer:

Parenchyma is the most simple and specialized tissue which is concerned mainly with the vegetative activities of the plant. The cells are isodiametric with well-developed intercellular spaces, vacuolated cytoplasm and cellulosic cell wall.

Collenchyma is the tissue of the primary body. The cells of the tissue contain protoplasm and are living without intercellular spaces. The cell wall articulate at the corners and are made up of cellulose, hemicellulose, and pectin.

Sclerenchyma is the thick-walled cell tissue. In the beginning, the cell is living and have protoplasm, but due to deposition of impermeable secondary board lignin, they become dead thick and hard.

5 0

3 years ago

Which event must always occur for a chemical reaction to occur

Lorico [155]

There must be effective collisions between the reacting chemical particles in order for the chemical reaction to occur.

8 0

3 years ago

The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3)

cestrela7 [59]

Answer:

a. The limiting reactant is $NaHCO_{3}$

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

$3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}$ ⇒ $3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}$

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

Na: 23
H: 1
C: 12
O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

$NaHCO_{3}$ :23+1+12+16*3=84 g/mol
$H_{3} C_{6} HO_{7}$ :1*3+12*6+1*5+16*7= 192 g/mol
$CO_{2}$ :12+16*2= 44 g/mol
$H_{2} O$ :1*2+16= 18 g/mol
$Na_{3} C_{6} H_{5} O_{7}$ : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of $NaHCO_{3}$ react with 1 mole of $H_{3} C_{6} HO_{7}$ Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

$grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}$

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So the limiting reagent is sodium bicarbonate.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of $NaHCO_{3}$ react with 3 mole of $CO_{2}$ . Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

$NaHCO_{3}$ : 252 g
$H_{3} C_{6} HO_{7}$ : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

$grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}$

grams of carbon dioxide= 0.73 g

Then, 0.73 g of carbon dioxide are formed.

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

$grams of citric acid=\frac{1.4 g * 192 g}{252 g}$

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

So the grams of excess reactant that do not participate in the reaction are 0333 g.

3 0

3 years ago